Question: The polar curve $r(\theta)=2-4\sin(\theta)$ is graphed for $0\leq\theta\leq\pi$. Let $R$ be the region in the third and fourth quadrants enclosed by the curve, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{\pi}{8}}^{\scriptsize\dfrac{7\pi}{8}}\left( 1-4\sin(\theta)+4\sin^2(\theta)\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{5\pi}{6}}\left( 1-4\sin(\theta)+4\sin^2(\theta)\right)d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{5\pi}{6}}\left( 2-8\sin(\theta)+8\sin^2(\theta)\right)d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{\pi}{8}}^{\scriptsize\dfrac{7\pi}{8}}\left( 2-8\sin(\theta)+8\sin^2(\theta)\right)d\theta$
Answer: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. Notice that the curve goes through the third and fourth quadrants, even though it's graphed for $0\leq\theta\leq\pi$. That's because for a certain range of $\theta$ -values, the sign of $r(\theta)$ is negative. The part of the curve that encloses $R$ starts at a point where $r(\theta)=0$ and ends at the next point where $r(\theta)=0$. The $\theta$ -values in the range $0\leq\theta\leq\pi$ for which $r(\theta)=0$ are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$. So $\alpha$ must be $\dfrac{\pi}{6}$ and $\beta$ must be $\dfrac{5\pi}{6}$. Let's plug ${r(\theta)=2-4\sin(\theta)}$, ${\alpha=\dfrac{\pi}{6}}$, and ${\beta=\dfrac{5\pi}{6}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{6}}}^{{\scriptsize\dfrac{5\pi}{6}}}\dfrac{1}{2}\left({2-4\sin(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{5\pi}{6}}\dfrac{1}{2}\left( 4-16\sin(\theta)+16\sin^2(\theta)\right)d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{5\pi}{6}}\left( 2-8\sin(\theta)+8\sin^2(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{6}}^{\scriptsize\dfrac{5\pi}{6}}\left( 2-8\sin(\theta)+8\sin^2(\theta)\right)d\theta$